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Number Representations
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Numbers with Fractions
So, done with negative numbers. Done with signed and unsigned integers. How about numbers with fractions? How to represent, say, 5.75ten in binary forms?
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Numbers with Fractions
In general, to represent a real number in binary, you first find the binary representation of the integer part, then find the binary representation of the fraction part, then put a dot in between.
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Numbers with fractions
The integer part is 5ten which is 101two. How did you get it?
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Numbers with Fractions
The fraction is Note that it is = , so 5.75ten = two
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How to get the fraction In general, what you do is kind of the reverse of getting the binary representation for the integer: divide the fraction first by 0.5 (2-1), take the quotient as the first bit of the binary fraction, then divide the remainder by 0.25 (2-2), take the quotient as the second bit of the binary fraction, then divide the remainder by (2-3),
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How to get the fraction Take 0.1 as an example.
0.1/0.5=0* –> bit 1 is 0. 0.1/0.25 = 0* –> bit 2 is 0. 0.1/0.125 = 0* –> bit 3 is 0. 0.1/ = 1* –> bit 4 is 1. 0.0375/ = 1* –> bit 5 is 1. And so on, until the you have used all the bits that hardware permits
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Floating Point Numbers
11/19/2018 Floating Point Numbers Recall scientific notation for decimal numbers A number is represented by a significand (or mantissa) and an integer exponent F * 10E Where F is the significand, and E the exponent Example: * 102 Normalized if F is a single digit number 11/19/2018 CDA3100 CDA3100
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Floating Points in Binary
11/19/2018 Floating Points in Binary Normalized binary scientific notation For a fixed number of bits, we need to decide How many bits for the significand (or fraction) How many bits for the exponent There is a trade-off between precision and range More bits for significand increases precision while more bits for exponent increases the range 11/19/2018 CDA3100 CDA3100
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IEEE 754 Floating Point Standard
11/19/2018 IEEE 754 Floating Point Standard Single precision Represented by 32 bits Since the leading 1 bit in the significand in normalized binary numbers is always 1, it is not represented explicitly 11/19/2018 CDA3100 CDA3100
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11/19/2018 Exponent If we represent exponents using two’s complement, then it would not be intuitive as small numbers appear to be larger 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 11/19/2018 CDA3100 CDA3100
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11/19/2018 Biased Notation The most negative exponent will be represented as 00…00 and the most positive as 111…11 That is, we need to subtract the bias from the corresponding unassigned value The value of an IEEE 754 single precision is 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 s exponent fraction 1 bit 8 bits 23 bits 11/19/2018 CDA3100 CDA3100
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Example 101.11two= 22 + 20 + 2-1 + 2-2 = 5.75ten
11/19/2018 Example 101.11two= = 5.75ten The normalized binary number will be 22 = 2( ) So the exponent is 129ten = As a hexadecimal number, the representation is 0x40B80000 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 11/19/2018 CDA3100 CDA3100
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IEEE 754 Double Precision It uses 64 bits (two 32-bit words)
11/19/2018 IEEE 754 Double Precision It uses 64 bits (two 32-bit words) 1 bit for the sign 11 bits for the exponent 52 bits for the fraction 1023 as the bias 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 s Exponent fraction 1 bit 11 bits 20 bits Fraction (continued) 32 bits 11/19/2018 CDA3100 CDA3100
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Example (Double Precision)
11/19/2018 Example (Double Precision) 101.11two= = 5.75 The normalized binary number will be 22 = 2( ) So the exponent is 1025ten = two As a hexadecimal number, the representation is 0x 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 11/19/2018 CDA3100 CDA3100
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floating-point number
11/19/2018 Special Cases Single precision Double precision Object represented Exponent Fraction nonzero denormalized number 1-254 anything 1-2046 floating-point number 255 2047 infinity NaN (Not a number) Denormalized If the exponent is all 0s, but the fraction is non-zero (else it would be interpreted as zero), then the value is a denormalized number, which does not have an assumed leading 1 before the binary point. Thus, this represents a number (-1)s × 0.f × 2-126, where s is the sign bit and f is the fraction. For double precision, denormalized numbers are of the form (-1)s × 0.f × From this you can interpret zero as a special type of denormalized number. 11/19/2018 CDA3100 CDA3100
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Floating Point Numbers
How many different numbers can the single precision format represent? What is the largest number it can represent?
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Ranges for IEEE 754 Single Precision
11/19/2018 Ranges for IEEE 754 Single Precision Largest positive number Smallest positive number (floating point) 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 11/19/2018 CDA3100 CDA3100
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Ranges for IEEE 754 Single Precision
11/19/2018 Ranges for IEEE 754 Single Precision Largest positive number Smallest positive number (floating point) 11/19/2018 CDA3100 CDA3100
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Ranges for IEEE 754 Double Precision
11/19/2018 Ranges for IEEE 754 Double Precision Largest positive number Smallest positive number (Floating-point number) 11/19/2018 CDA3100 CDA3100
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Comments on Overflow and Underflow
11/19/2018 Comments on Overflow and Underflow Overflow (and underflow also for floating numbers) happens when a number is outside the range of a particular representation For example, by using 8-bit two’s complement representation, we can only represent a number between and 127 If a number is smaller than -128, it will cause overflow If a number is larger than 127, it will cause overflow also Note that arithmetic operations can result in overflow 11/19/2018 CDA3100 CDA3100
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